Question

Two pipes running together can fill a cistern in \[3\dfrac{1}{13}\] minutes. If one pipe takes 3 minutes more than the other to fill it, find the time in which each pipe would fill the cistern.

Answer 1

Hint: Assume the volume of cistern as ‘V’ and the time taken by pipe 1 to fill the cistern alone as ‘t’. Using this assumption find the time taken by pipe 2 to fill the cistern alone. Now, use the unitary method to find the volume of cistern filled by each pipe alone in 1 minute and take their sum and substitute it with the volume of cistern filled by both the piper together in a minute. Form a quadratic equation in ‘t’ and apply the middle term split method to find the value of ‘t’

Complete step by step answer:
Let us assume the volume of cistern as ‘V’. Now, we are assuming that pipe 1 takes time ‘t’ to fill the cistern alone.
Here, it is given that the pipe 2 takes minutes more to fill the cistern alone. So, the time taken by pipe 2 to fill the cistern alone will be (t + 3) minutes.
Now, volume of cistern filled by pipe 1 in t minutes = V.
\[\Rightarrow \] Volume of cistern filled by pipe 1 in 1 minute = \[\dfrac{V}{t}\].
Similarly, volume of cistern filled by pipe 2 in (t + 3) minutes = V
\[\Rightarrow \] Volume of cistern filled by pipe 2 in 1 minute = \[\dfrac{V}{t+3}\].
So, we have total volume of cistern filled by pipe 1 and pipe 2 together in 1 minute = \[\dfrac{V}{t}+\dfrac{V}{t+3}\] - (1)
Now, it is given that both the pipes together fill the cistern in \[3\dfrac{1}{13}\] minutes. Converting this mixed fraction into improper fraction, we get,
\[\Rightarrow \]\[3\dfrac{1}{13}\] minutes \[=\dfrac{39+1}{13}=\dfrac{40}{13}\] minutes
\[\therefore \] Volume of cistern filled by both the pipes in \[\dfrac{40}{13}\] minutes = V
\[\Rightarrow \] Volume of cistern filled by both the pipes in 1 minute = \[\dfrac{V}{\dfrac{40}{13}}\]
\[\Rightarrow \] Volume of cistern filled by both the pipes in 1 minute = \[\dfrac{13V}{40}\] - (2)
Comparing equations (1) and (2), we get,
\[\Rightarrow \dfrac{V}{t}+\dfrac{V}{t+3}=\dfrac{13V}{40}\]
Cancelling ‘V’ from all the terms, we get,
\[\begin{align}
  & \Rightarrow \dfrac{1}{t}+\dfrac{1}{t+3}=\dfrac{13}{40} \\
 & \Rightarrow \dfrac{t+3+t}{{{t}^{2}}+3t}=\dfrac{13}{40} \\
\end{align}\]
By cross – multiplication, we get,
\[\begin{align}
  & \Rightarrow 80t+120=13{{t}^{2}}+39t \\
 & \Rightarrow 13{{t}^{2}}-41t-120=0 \\
\end{align}\]
Using the splitting of middle term method, we get,
\[\begin{align}
  & \Rightarrow 13{{t}^{2}}-65t+24t-120=0 \\
 & \Rightarrow 13t\left( t-5 \right)+24\left( t-5 \right)=0 \\
 & \Rightarrow \left( t-5 \right)\left( 13t+24 \right)=0 \\
\end{align}\]
\[\Rightarrow t=5\] or \[\dfrac{-24}{13}\]
Since, time cannot be negative, therefore \[t=\dfrac{-24}{13}\] is neglected.
\[\Rightarrow t=5\] minutes
\[\Rightarrow \left( t+3 \right)=8\] minutes

Hence, time taken by pipe 1 and pipe 2 to fill the cistern alone are 5 minutes and 8 minutes respectively.

Note: One may note that we have assumed the time taken by pipe 1 as ‘t’ and time taken by pipe 2 as ‘t + 3’. You can also assume the time taken as ‘t’ for pipe 1 and ‘t - 3’ for pipe 2. This will not change the numerical values but only the time taken by both the pipes will get interchanged. Remember that the quadratic equation will give two values of t in which the negative value must be rejected because time is not negative.